Question

(II) What is the minimum work needed to push a 950-kg car 710 m up along a 9.0° incline? Ignore friction?

Short answer

The obtained value of work done is \(1.0 \times {10^6}\;{\rm{J}}\).

Step-by-step solution

Determine the forces in the horizontal direction

Draw the free body diagram of the car moving up along an inclined plane. Consider all the forces acting on the car in the horizontal and vertical directions.

The minimum work is done on the car when it is moving with steady velocity.

Given data:

The mass of the car is\(m = 950\;{\rm{kg}}\).

The distance is\(d = 710\;{\rm{m}}\).

The incline angle is\(\theta = 9^\circ \).

The free body diagram of the car is as follows:

Consider the x-direction along the plane and the y-direction perpendicular to the plane. The relation of the forces in the x-direction is given by:

\(\begin{aligned}\Sigma {F_x} &= 0\\F - W\sin \theta &= 0\\F &= mg\sin \theta \end{aligned}\)

Here, F is the pushing force, W is the weight of the car, and g is the gravitational acceleration.

Determine the minimum work done to push the car

The relation of work done for the displacement in the direction of the force is given by:

\(W = F \times d\)

On plugging the values in the above relation, you get:

\(\begin{aligned}W &= mg\sin \theta \times d\\W &= \left( {950\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\sin 9^\circ \left( {710\;{\rm{m}}} \right)\\W &= 10.35 \times {10^5}\;{\rm{J}}\\W \approx 1.0 \times {10^6}{\rm{ J}}\end{aligned}\)

Thus, \(W = 1.0 \times {10^6}\;{\rm{J}}\) is the required work done.